f(x)是二次函数,且f(x+4)+f(x-1)=x²-2x,求函数f(x）的解析式

设二次函数f(x)=ax^2+bx+c则a(x+4)^2+b(x+4)+c+a(x-1)^2+b(x-1)+c=x^2-2x化简,整理,得,2ax^2+(6a+2b)x+17a+3b+2c=x^2-2x系数待定法,得,2a=1,6a+2b=-2,17a+3b+2c=0得a=1/2,b=-5/2,c=-1/2所以,f(x)=1/2 x^2-5/2 x-1/2...